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प्रश्न
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
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उत्तर
LHS = `(1 - tanA)^2 + (1 + tanA)^2`
= `1 + tan^2A - 2tanA + 1 + tan^2A + 2tanA`
= `2(1 + tan^2A) = 2sec^2A` = RHS
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संबंधित प्रश्न
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
(i)` (1-cos^2 theta )cosec^2theta = 1`
`sqrt((1+sin theta)/(1-sin theta)) = (sec theta + tan theta)`
Write the value of `cosec^2 theta (1+ cos theta ) (1- cos theta).`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identities.
`(sin "A" - sin "B")/(cos "A" + cos "B") + (cos "A" - cos "B")/(sin "A" + sin "B")`
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
